To what volume should 10L of 0.5M CH3COOH (Ka = 1.8×10–5) be diluted in order to double the OH–

Question

concentration?

Answer 1

CH3COOH = HAc.
.....HAc --> H^+ + Ac^-
I....0.5.....0......0
C.....-x.....x......x
E....0.5-x...x......x

Ka = (H^+)(Ac^-)/(HAc)
Substitute the E line into Ka expression and solve for H^+ initially in the 10 L.
Then (OH^-) = Kw/(H^+) to calculate OH^-
Double that. Recalculate H^+. Plug that into the Ka expression for H^+ and Ac^- and solve for HAc.
Then L1 x M1 = L2 x M2. Substitute and solve for L2.
10L x 0.5M = ?L2 x M you found for HAc in the last calculation. It's late so you should confirm all of this but I ran the calculation and got about 40 L.

Did you look at the question I had about the titration of the weak acid with NaOH. I'm hoping you left something out of that post.

Answer 2

I checked and you did respond to my question; however, you didn't answer it.

Answer 3

Yes DrBob222...
Thanks for your response... I tried previous problem as given below.. Don't know whether correct...

Let 'a' be initial milli moles of HA and molarity of NaOH be 'x'
In both case solution is acidic.. ie acidic buffer

HA + NaOH --> NaA +H2O
a -- ---
-10x -10x ---
a-10x -10x 10x

Using Henderson eqn...
When 10 ml is added...
5 = PKa + log 10x/a-10x
When 20 ml is added...
5.6 = Pka+ log 20x/a-20x

Solving above eqns value of 'a' interms of 'x' can be obtained...
Substituting that in first eqn value of Pka was obtained as 5.3

Answer 4

I tried that early on. The problem I had is that we have two equations and three unknowns. The unknowns are a, x, and pKa. I don't know how you solve for a in terms of x without involving pKa. So I'm inclined to think your solution is not correct but I don't know how to correct it. This has stumped me.

Answer 5

5 = Pka+ log 10x/a-10x -----(1)
5.6 = Pka+ log 20x/a-20x -----(2)

(2) - (1) -->

0.6= log(20x/a-20x)- log(10x/a-10x)

Since Pka is cancelled during subtraction.

Solving the above eqn taking RHS in the form log a - log b = log a/b

We get a = 30x

Substituting this value in eqn (1)

5 =Pka + log 10x/20x

ie 5 = Pka + log 0.5

Solving.. Pka = 5.3

Answer 6

You're ok. I substituted the numbers into equation 2 (with pKa = 5.3 and pH = 5.6

Answer 7

Zn°(s)|Zn2+(aq),Cl−(aq)|AgCl(s)|Ag°(s)